Now, if he can't come up with a definition of
directional derivative, this is almost certainly nice, but it's wake, yes, you don't want to
calculate it all over again when you need to find the
directional derivative. So we need to find an
alternative way, an easier way. Maybe more computable way to find the directional
derivative in a given point. In order to do it, and then
I go into first of all remind ourselves some concepts
that we already know. First of all, let's
start with gradient, which is a vector of first
derivatives written out here. The scalar product of two
vectors which is a rule in our two-dimensional case is multiplication of
coordinates summed. Also, it has a geometrical
meaning which is basically multiplication of
extra stance towards the cosine of the
vector in between them, which is schematically drawn in the picture on the slide here. So first of all, let us revisit some basic
scalar product properties here is symmetrical with, well, no surprise here because obviously is
geometrical census symmetrical. You do not care on the order of multiplication
of two real numbers. For the others, it defines
the vector lengths. It is bilinear, and it defines the angle between vectors x, we
are rewriting new. It's the easiest part
of all our course. But what we also then know, we also do know what is
differential function. The differentiable function
is a function which is approximated by its
tangent plane thoroughly. In our case it, is infinitesimally
towards distance between a given point and
the point of approximation. So why do we building all of these stuff on
top of each other? Just to define directional
derivative more easily, let us just try to look
at our definition of differentiation in terms
of directional derivative. If we're computing the
actual derivative, then the change of
argument we've it earlier. It's t multiplied by x
coordinate of the direction, and the nature of y-coordinate is t multiplied by
y-coordinate of direction. More importantly, the distance between the function and the
approximation function at given point and the
approximation point is simply t since our
direction is normalized. So let us try to write
the definition of our directional derivative
down. What do we get? We get a limit when
t approaches zero, change of function divided by t. Assume that our function is differentiable
at the point a, b. Then the change of function
can be rewritten in terms of first derivative and
infinite testable function towards the distance. So let us write this
down and we are going to substitute the change
of argument with tl_x, and the change of y
argument as tl_y. Well, the actual distance between two dots with a t
as we computed total there. So what we do have here? We do have here sum in the denominator proportional
to the denominator, to the t. We can divide both nominator
and denominator by t, and thus, we get, well, the first term
which is independent from the value of the limit. As a second term, well, the same thing similarly. The last thing we have
here is small or from one, which is simply
infinitesimal function as you do remember. Thus, its limit is
just equals to zero. As a result, we get a
pretty nice notation. We get partial
derivative multiplied by the x coordinate of the direction plus partial
derivative targets y, multiplied by the y coordinate of the direction or
simply scalar product of gradient and our direction. That's extraordinarily
nice formula. So we established
an easy relation between the directional
derivative, hence, the gradient, and derivative itself with
a scalar product. So let us consider
some example here. Just let us take a look at function x squared
plus 2y squared. For example, at point 1, 2. Well, we are going to ask
ourselves a question, what is the direction with the maximal value of directional
derivative in each one? So what we need to do firstly, we do not argue even that, because the function f
is differentiable at the point m. Because
it's polynomial, thus, it has a decent
partial derivatives. We do remember our
sufficient condition of differentiability. So we need just to find here the gradient of the
function at the point 1, 2. In order to do so,
I am going just to write a partial derivative
towards x which is 2x, and partial derivative
towards y which is 4y. Then I'm going to substitute y and x with one and
two respectively. Thus, we get two, eight as a result,
that's our gradient. So what we are going
to think about here. We have some arbitrary direction cosine
Alpha, sine Alpha. We are going to speak about what its maximum value of
this very vivid expression. So firstly, we need to compute our directional derivative, which is simple scalar product of the written and
we've just found, and our arbitrary direction, which is two cosine, functions plus eight
sine out of Alpha, and which is nice
and pretty obvious. But it's tricky to answer what is the
maximum value and why. So in order to do it, we are going to draw some
pretty approximate picture. Assume that that's
our point 1, 2, doesn't look like
it, but never mind. We have here our gradient. So we're talking
about is computing our directional derivative
with arbitrary direction l. Let us revisit the idea
what scalar products. What we do have here? We do have here also not only simple formula
for coordinates, but also eight geometrical
mean and [inaudible]. We can state that it's the length of our gradient multiplied
by the length of l, multiply it by some angle. We are going to call
it, for example, Gamma. Well, the length of the gradient. Well, it's quite easy. It is two squared plus a squared. That's not square. It's
the square root of 68. Let it be. The length of the
direction is always one, and this thing that
is left is, well, cosine of some
arbitrary angle Gamma. Thus, in order to come up
with the maximum value, you just need to
understand what's the maximum value of cosine
function which is one. So the maximum here
is square root of 68. By doing so, we've actually come to very interesting idea here. Because not only we spoke
about whether or not in geometrical meaning
is more important than the actual computation of rule for the
scalar product rule, we also found that the maximum value of
directional derivative somehow equals to the length
of the gradient. We're going to use that rule for sine in
order to establish the direction of maximum
gross and the speed of maximal gross of the
function at a given point.